Evaluating the limit using Taylor Series
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We're asked to find the following limit by using Taylor expansions $$lim_{xto{}0}frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}$$
My Attempt:
Expressing $e^{3x}$, $sin(x)$, $cos(x)$, $ln(1-2x)$ and $cos(5x)$ in their respective taylor expansions yielded the following monstrous fraction, https://imgur.com/a/xGyfIyL (Picture size too big to be uploaded here for some reason, plus fraction too large to be expressed in the space given :/) But anyways, I can't seem to factorize this thing and evaluate the limit as $xto{}0$, any help would be appreciated.
calculus limits analysis taylor-expansion
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up vote
3
down vote
favorite
We're asked to find the following limit by using Taylor expansions $$lim_{xto{}0}frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}$$
My Attempt:
Expressing $e^{3x}$, $sin(x)$, $cos(x)$, $ln(1-2x)$ and $cos(5x)$ in their respective taylor expansions yielded the following monstrous fraction, https://imgur.com/a/xGyfIyL (Picture size too big to be uploaded here for some reason, plus fraction too large to be expressed in the space given :/) But anyways, I can't seem to factorize this thing and evaluate the limit as $xto{}0$, any help would be appreciated.
calculus limits analysis taylor-expansion
A hint is that only the leading terms are going to matter. Can you see what happens to the constant, $x$, and $x^2$ terms in the numerator and denominator?
– user113102
yesterday
As @gimusi notes, you do not need the explicit Taylor series for the fraction. You can answer the question with some algebra on the leading terms for numerator and denominator.
– Ethan Bolker
yesterday
The main point is to guess the correct order for the expansion. In the doubt we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$. When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess looking at the denominator.
– gimusi
yesterday
I see now how I can go about evaluating the limit itself although I still find the concept a little bit vague, as in considering a specific order for the expansion and then applying it for all the functions while leaving the rest of the series behind, it'll probably start making more sense in a few more examples. Thanks for the help :)
– kareem bokai
yesterday
@kareembokai Yes of course you need only to practice to get confident with the method. The main rule is that all the function must be expanded to the same order. Note also that we can get an uncorrect evaluation when the order is too low than the minimum requested but not when the order is higher (but it requires more effort and work). Then you need to practice a lot with that in order to becomemore and more confident.
– gimusi
yesterday
|
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
We're asked to find the following limit by using Taylor expansions $$lim_{xto{}0}frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}$$
My Attempt:
Expressing $e^{3x}$, $sin(x)$, $cos(x)$, $ln(1-2x)$ and $cos(5x)$ in their respective taylor expansions yielded the following monstrous fraction, https://imgur.com/a/xGyfIyL (Picture size too big to be uploaded here for some reason, plus fraction too large to be expressed in the space given :/) But anyways, I can't seem to factorize this thing and evaluate the limit as $xto{}0$, any help would be appreciated.
calculus limits analysis taylor-expansion
We're asked to find the following limit by using Taylor expansions $$lim_{xto{}0}frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}$$
My Attempt:
Expressing $e^{3x}$, $sin(x)$, $cos(x)$, $ln(1-2x)$ and $cos(5x)$ in their respective taylor expansions yielded the following monstrous fraction, https://imgur.com/a/xGyfIyL (Picture size too big to be uploaded here for some reason, plus fraction too large to be expressed in the space given :/) But anyways, I can't seem to factorize this thing and evaluate the limit as $xto{}0$, any help would be appreciated.
calculus limits analysis taylor-expansion
calculus limits analysis taylor-expansion
asked yesterday
kareem bokai
324
324
A hint is that only the leading terms are going to matter. Can you see what happens to the constant, $x$, and $x^2$ terms in the numerator and denominator?
– user113102
yesterday
As @gimusi notes, you do not need the explicit Taylor series for the fraction. You can answer the question with some algebra on the leading terms for numerator and denominator.
– Ethan Bolker
yesterday
The main point is to guess the correct order for the expansion. In the doubt we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$. When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess looking at the denominator.
– gimusi
yesterday
I see now how I can go about evaluating the limit itself although I still find the concept a little bit vague, as in considering a specific order for the expansion and then applying it for all the functions while leaving the rest of the series behind, it'll probably start making more sense in a few more examples. Thanks for the help :)
– kareem bokai
yesterday
@kareembokai Yes of course you need only to practice to get confident with the method. The main rule is that all the function must be expanded to the same order. Note also that we can get an uncorrect evaluation when the order is too low than the minimum requested but not when the order is higher (but it requires more effort and work). Then you need to practice a lot with that in order to becomemore and more confident.
– gimusi
yesterday
|
show 1 more comment
A hint is that only the leading terms are going to matter. Can you see what happens to the constant, $x$, and $x^2$ terms in the numerator and denominator?
– user113102
yesterday
As @gimusi notes, you do not need the explicit Taylor series for the fraction. You can answer the question with some algebra on the leading terms for numerator and denominator.
– Ethan Bolker
yesterday
The main point is to guess the correct order for the expansion. In the doubt we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$. When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess looking at the denominator.
– gimusi
yesterday
I see now how I can go about evaluating the limit itself although I still find the concept a little bit vague, as in considering a specific order for the expansion and then applying it for all the functions while leaving the rest of the series behind, it'll probably start making more sense in a few more examples. Thanks for the help :)
– kareem bokai
yesterday
@kareembokai Yes of course you need only to practice to get confident with the method. The main rule is that all the function must be expanded to the same order. Note also that we can get an uncorrect evaluation when the order is too low than the minimum requested but not when the order is higher (but it requires more effort and work). Then you need to practice a lot with that in order to becomemore and more confident.
– gimusi
yesterday
A hint is that only the leading terms are going to matter. Can you see what happens to the constant, $x$, and $x^2$ terms in the numerator and denominator?
– user113102
yesterday
A hint is that only the leading terms are going to matter. Can you see what happens to the constant, $x$, and $x^2$ terms in the numerator and denominator?
– user113102
yesterday
As @gimusi notes, you do not need the explicit Taylor series for the fraction. You can answer the question with some algebra on the leading terms for numerator and denominator.
– Ethan Bolker
yesterday
As @gimusi notes, you do not need the explicit Taylor series for the fraction. You can answer the question with some algebra on the leading terms for numerator and denominator.
– Ethan Bolker
yesterday
The main point is to guess the correct order for the expansion. In the doubt we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$. When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess looking at the denominator.
– gimusi
yesterday
The main point is to guess the correct order for the expansion. In the doubt we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$. When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess looking at the denominator.
– gimusi
yesterday
I see now how I can go about evaluating the limit itself although I still find the concept a little bit vague, as in considering a specific order for the expansion and then applying it for all the functions while leaving the rest of the series behind, it'll probably start making more sense in a few more examples. Thanks for the help :)
– kareem bokai
yesterday
I see now how I can go about evaluating the limit itself although I still find the concept a little bit vague, as in considering a specific order for the expansion and then applying it for all the functions while leaving the rest of the series behind, it'll probably start making more sense in a few more examples. Thanks for the help :)
– kareem bokai
yesterday
@kareembokai Yes of course you need only to practice to get confident with the method. The main rule is that all the function must be expanded to the same order. Note also that we can get an uncorrect evaluation when the order is too low than the minimum requested but not when the order is higher (but it requires more effort and work). Then you need to practice a lot with that in order to becomemore and more confident.
– gimusi
yesterday
@kareembokai Yes of course you need only to practice to get confident with the method. The main rule is that all the function must be expanded to the same order. Note also that we can get an uncorrect evaluation when the order is too low than the minimum requested but not when the order is higher (but it requires more effort and work). Then you need to practice a lot with that in order to becomemore and more confident.
– gimusi
yesterday
|
show 1 more comment
4 Answers
4
active
oldest
votes
up vote
5
down vote
accepted
HINT
By Taylor's expansion, term by term, we have that
- $e^{3x}=1+3x+frac92x^2+o(x^2)$
- $sin x =x+o(x^2)$
- $cos x = 1-frac12 x^2+o(x^2)$
- $log(1-2x)=-2x-2x^2+o(x^2)$
- $cos (5x) = 1-frac{25}2 x^2+o(x^2)$
and then
$$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x+frac92x^2-x-1+frac12x^2-2x-2x^2+o(x^2)}{-1+1-frac{25}2x^2+o(x^2)}$$
Can you conclude from here?
Edit for a remark
The main point with Taylor's expansion is to guess the correct order to use for the expansion and there is not general a rule to be sure about the order to use.
In the doubt, we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$.
When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess, in that case, looking at the denominator which is in the form $cx^2+o(x^2)$.
I do see that you've ended up with the leading terms, however I still don't see how you managed to get rid of all the other terms so quickly
– kareem bokai
yesterday
@kareembokai I add something on that!
– gimusi
yesterday
@kareembokai Let me know whether now it is clear or not. In case do not hesitate to ask for any clarification.
– gimusi
yesterday
In general it's better to start with the term that appears simpler, in this case the denominator, which is easily seen to have order $2$.
– egreg
yesterday
@egreg Yes I agree in that case is quite simple, is some other cases for a student new in the topic it can be not so straightforward to see, but after a while and some tons of limits it becomes a simple task for usual limits.
– gimusi
yesterday
add a comment |
up vote
2
down vote
There is nothing monstrous in that computation. If you want a compact resolution, compute separately the coefficients in increasing order (numerator/denominator).
constant terms: $1-1=0 / -1+1=0$;
linear terms: $3-1-2=0 / 0$;
quadratic terms: $dfrac92+dfrac12-2=dfrac{6}{2} / -dfrac{25}2$.
As the first nonzero coefficients are of the same order, the limit is finite and is the ratio
$$-frac{6}{25}.$$
The trick is to obtain a fraction like
$$frac{ax^n+text{higher order terms}}{bx^m+text{higher order terms}}=x^{n-m}frac{a+text{higher order terms}}{b+text{higher order terms}}$$ which tends to $0,dfrac ab$ or $pminfty$ depending on the sign of $n-m$.
1
Nice way to simplify even if it requires a few of experience to be handle in that way.
– gimusi
yesterday
add a comment |
up vote
2
down vote
From your picture, it seems an expansion at order $2$ will do. All functions should be expanded at the same
order, using, say Taylor-Young's formula. Thus
$mathrm e^{3x}=1+3x+frac92x^2+o(x^2)$,
$sin x =x+o(x^2)$,
$cos x=1-frac12 x^2+o(x^2)$,
$ln(1-2x)=-2x-frac42 x^2+o(x^2)$
Thus the numerator is
$$N(x)=1+3x+frac92x^2-x-1+frac12 x^2-2x-2 x^2+o(x^2)=3 x^2+o(x^2).$$
Can you proceed?
add a comment |
up vote
2
down vote
We need only quote the numerator and denominator up to $x^2$ terms: $$lim_{xto 0}frac{1+3x+color{blue}{9x^2/2}-x-1+color{blue}{x^2/2}-2xcolor{blue}{-2x^2}+O(x^3)}{-1+1color{blue}{-25x^2/2}+O(x^3)}.$$You'll find only $x^2$ terms survive in each.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
HINT
By Taylor's expansion, term by term, we have that
- $e^{3x}=1+3x+frac92x^2+o(x^2)$
- $sin x =x+o(x^2)$
- $cos x = 1-frac12 x^2+o(x^2)$
- $log(1-2x)=-2x-2x^2+o(x^2)$
- $cos (5x) = 1-frac{25}2 x^2+o(x^2)$
and then
$$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x+frac92x^2-x-1+frac12x^2-2x-2x^2+o(x^2)}{-1+1-frac{25}2x^2+o(x^2)}$$
Can you conclude from here?
Edit for a remark
The main point with Taylor's expansion is to guess the correct order to use for the expansion and there is not general a rule to be sure about the order to use.
In the doubt, we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$.
When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess, in that case, looking at the denominator which is in the form $cx^2+o(x^2)$.
I do see that you've ended up with the leading terms, however I still don't see how you managed to get rid of all the other terms so quickly
– kareem bokai
yesterday
@kareembokai I add something on that!
– gimusi
yesterday
@kareembokai Let me know whether now it is clear or not. In case do not hesitate to ask for any clarification.
– gimusi
yesterday
In general it's better to start with the term that appears simpler, in this case the denominator, which is easily seen to have order $2$.
– egreg
yesterday
@egreg Yes I agree in that case is quite simple, is some other cases for a student new in the topic it can be not so straightforward to see, but after a while and some tons of limits it becomes a simple task for usual limits.
– gimusi
yesterday
add a comment |
up vote
5
down vote
accepted
HINT
By Taylor's expansion, term by term, we have that
- $e^{3x}=1+3x+frac92x^2+o(x^2)$
- $sin x =x+o(x^2)$
- $cos x = 1-frac12 x^2+o(x^2)$
- $log(1-2x)=-2x-2x^2+o(x^2)$
- $cos (5x) = 1-frac{25}2 x^2+o(x^2)$
and then
$$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x+frac92x^2-x-1+frac12x^2-2x-2x^2+o(x^2)}{-1+1-frac{25}2x^2+o(x^2)}$$
Can you conclude from here?
Edit for a remark
The main point with Taylor's expansion is to guess the correct order to use for the expansion and there is not general a rule to be sure about the order to use.
In the doubt, we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$.
When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess, in that case, looking at the denominator which is in the form $cx^2+o(x^2)$.
I do see that you've ended up with the leading terms, however I still don't see how you managed to get rid of all the other terms so quickly
– kareem bokai
yesterday
@kareembokai I add something on that!
– gimusi
yesterday
@kareembokai Let me know whether now it is clear or not. In case do not hesitate to ask for any clarification.
– gimusi
yesterday
In general it's better to start with the term that appears simpler, in this case the denominator, which is easily seen to have order $2$.
– egreg
yesterday
@egreg Yes I agree in that case is quite simple, is some other cases for a student new in the topic it can be not so straightforward to see, but after a while and some tons of limits it becomes a simple task for usual limits.
– gimusi
yesterday
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
HINT
By Taylor's expansion, term by term, we have that
- $e^{3x}=1+3x+frac92x^2+o(x^2)$
- $sin x =x+o(x^2)$
- $cos x = 1-frac12 x^2+o(x^2)$
- $log(1-2x)=-2x-2x^2+o(x^2)$
- $cos (5x) = 1-frac{25}2 x^2+o(x^2)$
and then
$$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x+frac92x^2-x-1+frac12x^2-2x-2x^2+o(x^2)}{-1+1-frac{25}2x^2+o(x^2)}$$
Can you conclude from here?
Edit for a remark
The main point with Taylor's expansion is to guess the correct order to use for the expansion and there is not general a rule to be sure about the order to use.
In the doubt, we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$.
When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess, in that case, looking at the denominator which is in the form $cx^2+o(x^2)$.
HINT
By Taylor's expansion, term by term, we have that
- $e^{3x}=1+3x+frac92x^2+o(x^2)$
- $sin x =x+o(x^2)$
- $cos x = 1-frac12 x^2+o(x^2)$
- $log(1-2x)=-2x-2x^2+o(x^2)$
- $cos (5x) = 1-frac{25}2 x^2+o(x^2)$
and then
$$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x+frac92x^2-x-1+frac12x^2-2x-2x^2+o(x^2)}{-1+1-frac{25}2x^2+o(x^2)}$$
Can you conclude from here?
Edit for a remark
The main point with Taylor's expansion is to guess the correct order to use for the expansion and there is not general a rule to be sure about the order to use.
In the doubt, we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$.
When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess, in that case, looking at the denominator which is in the form $cx^2+o(x^2)$.
edited yesterday
answered yesterday
gimusi
90.7k74495
90.7k74495
I do see that you've ended up with the leading terms, however I still don't see how you managed to get rid of all the other terms so quickly
– kareem bokai
yesterday
@kareembokai I add something on that!
– gimusi
yesterday
@kareembokai Let me know whether now it is clear or not. In case do not hesitate to ask for any clarification.
– gimusi
yesterday
In general it's better to start with the term that appears simpler, in this case the denominator, which is easily seen to have order $2$.
– egreg
yesterday
@egreg Yes I agree in that case is quite simple, is some other cases for a student new in the topic it can be not so straightforward to see, but after a while and some tons of limits it becomes a simple task for usual limits.
– gimusi
yesterday
add a comment |
I do see that you've ended up with the leading terms, however I still don't see how you managed to get rid of all the other terms so quickly
– kareem bokai
yesterday
@kareembokai I add something on that!
– gimusi
yesterday
@kareembokai Let me know whether now it is clear or not. In case do not hesitate to ask for any clarification.
– gimusi
yesterday
In general it's better to start with the term that appears simpler, in this case the denominator, which is easily seen to have order $2$.
– egreg
yesterday
@egreg Yes I agree in that case is quite simple, is some other cases for a student new in the topic it can be not so straightforward to see, but after a while and some tons of limits it becomes a simple task for usual limits.
– gimusi
yesterday
I do see that you've ended up with the leading terms, however I still don't see how you managed to get rid of all the other terms so quickly
– kareem bokai
yesterday
I do see that you've ended up with the leading terms, however I still don't see how you managed to get rid of all the other terms so quickly
– kareem bokai
yesterday
@kareembokai I add something on that!
– gimusi
yesterday
@kareembokai I add something on that!
– gimusi
yesterday
@kareembokai Let me know whether now it is clear or not. In case do not hesitate to ask for any clarification.
– gimusi
yesterday
@kareembokai Let me know whether now it is clear or not. In case do not hesitate to ask for any clarification.
– gimusi
yesterday
In general it's better to start with the term that appears simpler, in this case the denominator, which is easily seen to have order $2$.
– egreg
yesterday
In general it's better to start with the term that appears simpler, in this case the denominator, which is easily seen to have order $2$.
– egreg
yesterday
@egreg Yes I agree in that case is quite simple, is some other cases for a student new in the topic it can be not so straightforward to see, but after a while and some tons of limits it becomes a simple task for usual limits.
– gimusi
yesterday
@egreg Yes I agree in that case is quite simple, is some other cases for a student new in the topic it can be not so straightforward to see, but after a while and some tons of limits it becomes a simple task for usual limits.
– gimusi
yesterday
add a comment |
up vote
2
down vote
There is nothing monstrous in that computation. If you want a compact resolution, compute separately the coefficients in increasing order (numerator/denominator).
constant terms: $1-1=0 / -1+1=0$;
linear terms: $3-1-2=0 / 0$;
quadratic terms: $dfrac92+dfrac12-2=dfrac{6}{2} / -dfrac{25}2$.
As the first nonzero coefficients are of the same order, the limit is finite and is the ratio
$$-frac{6}{25}.$$
The trick is to obtain a fraction like
$$frac{ax^n+text{higher order terms}}{bx^m+text{higher order terms}}=x^{n-m}frac{a+text{higher order terms}}{b+text{higher order terms}}$$ which tends to $0,dfrac ab$ or $pminfty$ depending on the sign of $n-m$.
1
Nice way to simplify even if it requires a few of experience to be handle in that way.
– gimusi
yesterday
add a comment |
up vote
2
down vote
There is nothing monstrous in that computation. If you want a compact resolution, compute separately the coefficients in increasing order (numerator/denominator).
constant terms: $1-1=0 / -1+1=0$;
linear terms: $3-1-2=0 / 0$;
quadratic terms: $dfrac92+dfrac12-2=dfrac{6}{2} / -dfrac{25}2$.
As the first nonzero coefficients are of the same order, the limit is finite and is the ratio
$$-frac{6}{25}.$$
The trick is to obtain a fraction like
$$frac{ax^n+text{higher order terms}}{bx^m+text{higher order terms}}=x^{n-m}frac{a+text{higher order terms}}{b+text{higher order terms}}$$ which tends to $0,dfrac ab$ or $pminfty$ depending on the sign of $n-m$.
1
Nice way to simplify even if it requires a few of experience to be handle in that way.
– gimusi
yesterday
add a comment |
up vote
2
down vote
up vote
2
down vote
There is nothing monstrous in that computation. If you want a compact resolution, compute separately the coefficients in increasing order (numerator/denominator).
constant terms: $1-1=0 / -1+1=0$;
linear terms: $3-1-2=0 / 0$;
quadratic terms: $dfrac92+dfrac12-2=dfrac{6}{2} / -dfrac{25}2$.
As the first nonzero coefficients are of the same order, the limit is finite and is the ratio
$$-frac{6}{25}.$$
The trick is to obtain a fraction like
$$frac{ax^n+text{higher order terms}}{bx^m+text{higher order terms}}=x^{n-m}frac{a+text{higher order terms}}{b+text{higher order terms}}$$ which tends to $0,dfrac ab$ or $pminfty$ depending on the sign of $n-m$.
There is nothing monstrous in that computation. If you want a compact resolution, compute separately the coefficients in increasing order (numerator/denominator).
constant terms: $1-1=0 / -1+1=0$;
linear terms: $3-1-2=0 / 0$;
quadratic terms: $dfrac92+dfrac12-2=dfrac{6}{2} / -dfrac{25}2$.
As the first nonzero coefficients are of the same order, the limit is finite and is the ratio
$$-frac{6}{25}.$$
The trick is to obtain a fraction like
$$frac{ax^n+text{higher order terms}}{bx^m+text{higher order terms}}=x^{n-m}frac{a+text{higher order terms}}{b+text{higher order terms}}$$ which tends to $0,dfrac ab$ or $pminfty$ depending on the sign of $n-m$.
edited yesterday
answered yesterday
Yves Daoust
123k668219
123k668219
1
Nice way to simplify even if it requires a few of experience to be handle in that way.
– gimusi
yesterday
add a comment |
1
Nice way to simplify even if it requires a few of experience to be handle in that way.
– gimusi
yesterday
1
1
Nice way to simplify even if it requires a few of experience to be handle in that way.
– gimusi
yesterday
Nice way to simplify even if it requires a few of experience to be handle in that way.
– gimusi
yesterday
add a comment |
up vote
2
down vote
From your picture, it seems an expansion at order $2$ will do. All functions should be expanded at the same
order, using, say Taylor-Young's formula. Thus
$mathrm e^{3x}=1+3x+frac92x^2+o(x^2)$,
$sin x =x+o(x^2)$,
$cos x=1-frac12 x^2+o(x^2)$,
$ln(1-2x)=-2x-frac42 x^2+o(x^2)$
Thus the numerator is
$$N(x)=1+3x+frac92x^2-x-1+frac12 x^2-2x-2 x^2+o(x^2)=3 x^2+o(x^2).$$
Can you proceed?
add a comment |
up vote
2
down vote
From your picture, it seems an expansion at order $2$ will do. All functions should be expanded at the same
order, using, say Taylor-Young's formula. Thus
$mathrm e^{3x}=1+3x+frac92x^2+o(x^2)$,
$sin x =x+o(x^2)$,
$cos x=1-frac12 x^2+o(x^2)$,
$ln(1-2x)=-2x-frac42 x^2+o(x^2)$
Thus the numerator is
$$N(x)=1+3x+frac92x^2-x-1+frac12 x^2-2x-2 x^2+o(x^2)=3 x^2+o(x^2).$$
Can you proceed?
add a comment |
up vote
2
down vote
up vote
2
down vote
From your picture, it seems an expansion at order $2$ will do. All functions should be expanded at the same
order, using, say Taylor-Young's formula. Thus
$mathrm e^{3x}=1+3x+frac92x^2+o(x^2)$,
$sin x =x+o(x^2)$,
$cos x=1-frac12 x^2+o(x^2)$,
$ln(1-2x)=-2x-frac42 x^2+o(x^2)$
Thus the numerator is
$$N(x)=1+3x+frac92x^2-x-1+frac12 x^2-2x-2 x^2+o(x^2)=3 x^2+o(x^2).$$
Can you proceed?
From your picture, it seems an expansion at order $2$ will do. All functions should be expanded at the same
order, using, say Taylor-Young's formula. Thus
$mathrm e^{3x}=1+3x+frac92x^2+o(x^2)$,
$sin x =x+o(x^2)$,
$cos x=1-frac12 x^2+o(x^2)$,
$ln(1-2x)=-2x-frac42 x^2+o(x^2)$
Thus the numerator is
$$N(x)=1+3x+frac92x^2-x-1+frac12 x^2-2x-2 x^2+o(x^2)=3 x^2+o(x^2).$$
Can you proceed?
edited yesterday
answered yesterday
Bernard
117k637109
117k637109
add a comment |
add a comment |
up vote
2
down vote
We need only quote the numerator and denominator up to $x^2$ terms: $$lim_{xto 0}frac{1+3x+color{blue}{9x^2/2}-x-1+color{blue}{x^2/2}-2xcolor{blue}{-2x^2}+O(x^3)}{-1+1color{blue}{-25x^2/2}+O(x^3)}.$$You'll find only $x^2$ terms survive in each.
add a comment |
up vote
2
down vote
We need only quote the numerator and denominator up to $x^2$ terms: $$lim_{xto 0}frac{1+3x+color{blue}{9x^2/2}-x-1+color{blue}{x^2/2}-2xcolor{blue}{-2x^2}+O(x^3)}{-1+1color{blue}{-25x^2/2}+O(x^3)}.$$You'll find only $x^2$ terms survive in each.
add a comment |
up vote
2
down vote
up vote
2
down vote
We need only quote the numerator and denominator up to $x^2$ terms: $$lim_{xto 0}frac{1+3x+color{blue}{9x^2/2}-x-1+color{blue}{x^2/2}-2xcolor{blue}{-2x^2}+O(x^3)}{-1+1color{blue}{-25x^2/2}+O(x^3)}.$$You'll find only $x^2$ terms survive in each.
We need only quote the numerator and denominator up to $x^2$ terms: $$lim_{xto 0}frac{1+3x+color{blue}{9x^2/2}-x-1+color{blue}{x^2/2}-2xcolor{blue}{-2x^2}+O(x^3)}{-1+1color{blue}{-25x^2/2}+O(x^3)}.$$You'll find only $x^2$ terms survive in each.
edited yesterday
answered yesterday
J.G.
20.5k21933
20.5k21933
add a comment |
add a comment |
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A hint is that only the leading terms are going to matter. Can you see what happens to the constant, $x$, and $x^2$ terms in the numerator and denominator?
– user113102
yesterday
As @gimusi notes, you do not need the explicit Taylor series for the fraction. You can answer the question with some algebra on the leading terms for numerator and denominator.
– Ethan Bolker
yesterday
The main point is to guess the correct order for the expansion. In the doubt we could decide to start with the first order to obtain $$frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}=frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $frac 0 0$. When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess looking at the denominator.
– gimusi
yesterday
I see now how I can go about evaluating the limit itself although I still find the concept a little bit vague, as in considering a specific order for the expansion and then applying it for all the functions while leaving the rest of the series behind, it'll probably start making more sense in a few more examples. Thanks for the help :)
– kareem bokai
yesterday
@kareembokai Yes of course you need only to practice to get confident with the method. The main rule is that all the function must be expanded to the same order. Note also that we can get an uncorrect evaluation when the order is too low than the minimum requested but not when the order is higher (but it requires more effort and work). Then you need to practice a lot with that in order to becomemore and more confident.
– gimusi
yesterday